How can $10$ different pencils be distributed among $3$ students?
MY TRY $1$
total ways $= 3^{10}$
MY TRY $2$
$10 \times 9 \times 8 =720$
Which one is correct? If both are wrong what is correct answer? And please explain the approach.
EDIT
A query just came in my mind after the comment on this question… As he said $3^10$ is correct. Then my question is: if there were $n$ identical things (i.e. $10$ identical pencils) would the answer be same, $3^{10}$? Or would it be $\binom{n+r-1}{r-1}$ i.e. $\binom{12}2$ then?
Best Answer
As stated in comments the correct answer is $3^{10}$.
Now I will tell you what to do for identical pencils. You have to find how many ways can you distribute $10$ pencils to $3$ people. Let the number of pencils with each one be $a,b,c$
Clearly, $a+b+c=10$
Also $a,b,c$ are whole numbers. Now here is the trick.
Add 3 to both sides of equations and rename
$a+1=A$
$b+1=B$
$c+1=C$
This won't affect the number of your solutions. Hence, $A+B+C=13$ Why did I do so? See : You can write 13 as $1+1+1+1+1+1+1+1+1+1+1+1+1$
I'm not kidding. Now see left side. How many $+$ signs you see? And on the right?
Obviously, 2 on the left and 12 on the right. Now you just have to choose 2 plus signs on the right for that and the $1+1...$ which remain will form a number. Clearly, all these ways will be a solution to your problem. Marvelous, isn't it?
So the answer is $\binom{12}{2}$. No need to mug all those formulas...